Q 12-11-167JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300$ nm to $400$ nm. The decrease in the stopping potential is close to: $\left(\dfrac{hc}{e} = 1240\ \text{nm V}\right)$
Answer: (C) $1.0$ V
$eV_0 = \dfrac{hc}{\lambda} - \phi$, so
$$\Delta V_0 = 1240\left(\frac1{300} - \frac1{400}\right) = \frac{1240}{1200} \approx 1.0\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics